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A woman dropped a dime in a wishing well , she heard it 30 sec later ..Find velocity on impact while moving at 9.8 m/s 2

VMariaS [17]

Answer:

The velocity at impact is 222.2 m/s

Explanation:

The given information are;

The time at which the woman heard the sound of the coin after dropping it into the wishing well = 30 seconds

The average speed of sound in air = 344 m/s

The time in which the dime travel = t₁

The time in which the sound travel = t₂

1/2 × 9.8 × t₁² = 344 × t₂

t₁ + t₂ = 30

4.9·t₁² = 344 × (30 - t₁)

4.9·t₁² + 344·t₁ - 10320 = 0

Which gives -344±

$t_1 = \dfrac{-344\pm \sqrt{344^{2}-4\times 4.9 \times 10320}}{2\times 4.9}$

t₁ = 22.676 seconds or -92.9 seconds

Therefore the correct natural time is t₁ = 22.676 seconds

The velocity at impact, v = g×t₁ = 9.8 × 22.676 = 222.2 m/s

The velocity at impact= 222.2 m/s.

8 0

4 years ago

A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration

JulsSmile [24]

Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

$v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}$

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

$v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m$

So, the maximum height reached by the rocket is:

$h=y_1+y_2\\h=160m+154.28m=314.28m$

(c) In the first part we have:

$v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s$

And in the second part:

$t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s$

So, the time it takes to reach the maximum height is:

$t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s$

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

$v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}$

$t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s$

So, the total time is:

$t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s$

7 0

4 years ago

Una persona de 80 kilogramos de masa se encuentra parada sobre una silla de 3 kilogramos de masa Cuál es la presión que ejerce e

MariettaO [177]

Answer:

de 20 miligramos cada pata dividiendo 80÷4=20

7 0

3 years ago

A large group of people get together. each one rolls a die 180 times, and counts the number of 1's. about what percentage of the

Arte-miy333 [17]

<span>Around 99% - 100%. This is because a die is six sided, which gives the odds of a one coming up being roughly 17% independent of every roll. 17% of 180 trials comes out to 30-31 times a one will show up every 180 trials. This puts you right in the middle of the 15-45 range which means that somebody will almost ALWAYS reach 15-45 one's in a trial of 180 rolls.</span>

5 0

3 years ago

What happens to the volume of sound as the amplitude gets larger?

tino4ka555 [31]

The sound is perceived as louder if the amplifier increases, and softer if the amplitude decreases.

7 0

3 years ago

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